# (A^n+B^n)^(1/n) limit stuff

1. Jul 4, 2008

### jostpuur

If $$0<A<B$$ then

$$B < (A^n + B^n)^{1/n} < 2^{1/n} B,\quad\quad n=1,2,3,\ldots$$

and by using knowledge $$2^{1/n}\to 1$$ as $$n\to\infty$$, we get

$$\lim_{n\to\infty} (A^n + B^n)^{1/n} = B.$$

Now, I'm interested to know what can we say if $$-B<A<0<B$$. With odd n the inequality

$$B\leq (A^n + B^n)^{1/n}$$

is wrong, so there does not seem to be any reason to believe that the limit would still be B. But is the limit still B?

For example, set A=-1, B=2. The numbers

n, (-1)^n + 2^n, ((-1)^n + 2^n)^(1/n)

turn out as follows.

1, 1, 1
2, 5, 2.2361...
3, 7, 1.9129...
4, 17, 2.0305...
5, 31, 1.9873...
6, 65, 2.0052...
7, 127, 1.9978...

The 2^n appears to be dominating the limit quite well. Is the limit of this 2 really? What about more general conditions for A and B?

Last edited: Jul 4, 2008
2. Jul 4, 2008

### Werg22

I presume you meant |B| > A > 0 > B. (A^n + B^n)^1/n does not converge. This is best seen if you consider odd and even integers separately. If n only assumes even value, the expression converges to -B. If n assumes odd values, it converges to B. In a convergent sequence, every subsequence converges to the same value.

3. Jul 4, 2008

### jostpuur

Sorry. I fixed the mistake with edit option.

4. Jul 4, 2008

### jostpuur

I just realized it looks quite clear when you put it like this

$$(A^n + B^n)^{1/n} = B\underbrace{\Big(\underbrace{\big(\frac{A}{B}\big)^n}_{\to 0} + 1\Big)^{1/n}}_{\to 1} \to B,$$

but I still encountered some problems with the proof...

If we know that the sequence of functions

$$f_n:[1-\delta, 1+\delta]\to\mathbb{R},\quad f_n(x)=(1+x)^{1/n}$$

with fixed $$0<\delta <1$$, converges towards

$$f:[1-\delta, 1+\delta]\to\mathbb{R},\quad f(x)=1$$

uniformly, then the proof should be done, right? Looks clear by picture. I'm not sure how the uniform convergence is proven properly.

Last edited: Jul 4, 2008
5. Jul 4, 2008

### Werg22

Judging from what you were asking earlier, what you are implying is that B is negative (and so is A). If that's the case, factoring out B as you did is incorrect; (B^n)^1/n is not the same as B for an even n.

Yes, for every fixed x, f_n(x) does converge to 1 uniformly as n goes to infinity. However, we are interested in g_n(x) = K*f_n(x) where K = B if n is even and K = -B if n is odd. For any x in the said interval, g_n(x) does not converge as n goes to infinity (there is no limit).

Your argument works only in the case B is positive, in which case factoring out B is correct.

If I'm misunderstanding you, please pardon me.

6. Jul 4, 2008

### jostpuur

You must have been left confused by my original typo. Now the original post assumes -B < 0 < B, so B is positive.

With fixed x the word uniform doesn't have a meaning. What is relevant is that the convergence is uniform in some environment of x=1.

7. Jul 4, 2008

### Werg22

If B is positive, what is the question exactly?

By uniform convergence, I understood that you meant that for any x' in [1 - delta, 1 + delta], f_n(x') would form a uniformly decreasing sequence.

Last edited: Jul 4, 2008
8. Jul 4, 2008

### jostpuur

The questions is that is the limit

$$\lim_{n\to\infty}(A^n+B^n)^{1/n} = B$$

correct for all $$0<|A|<B$$. I already knew how to prove the case $$0<A<B$$, but the case $$-B<A<0<B$$ seemed to be more difficult. However, I believe I already understood how this is proven too, provided that we know the one uniform convergence result mentioned in the post #4.

I don't know what it means for a sequence of numbers to converge uniformly or not uniformly. The sequence of functions $$f_n:[a,b]\to\mathbb{R}$$ converges uniformly towards a function $$f:[a,b]\to\mathbb{R}$$, if for all $$\epsilon>0$$ there exists a number $$N\in\mathbb{N}$$ so that

$$n>N\quad\implies\quad \underset{x\in[a,b]}{\sup} |f_n(x)-f(x)| < \epsilon.$$

9. Jul 4, 2008

### jostpuur

... environment of x=0. The domains should have been $$[-\delta,\delta]$$ too, instead of $$[1-\delta, 1+\delta]$$.